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Let $a>0$. How to prove the following inequality $$\exists c>0,\exists A>0,\forall t>0:\quad\int^\infty_0 \frac{\sin(rt)}{rt}\frac{r^4}{\sinh^2(r)} e^{-ar\coth(r)}dr\leq c \big(e^{-At}\big)$$ for some constants $A>0,c>0$

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  • $\begingroup$ Does this mean $\forall t>0\,\,\exists A,c\,\,\cdots$ or $\exists A,c\,\,\forall t>0\,\,\cdots \text{?} \qquad$ $\endgroup$ Jul 8, 2022 at 19:00
  • $\begingroup$ It means $\exists A,c\,\,\forall t>0\,\,\cdots$. Thank you $\endgroup$ Jul 8, 2022 at 19:05
  • $\begingroup$ @ThomasKojar : I don't think that straightforward contour integration would work here, because the integrand is not meromorphic in $r$. $\endgroup$ Jul 10, 2022 at 2:41

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$\newcommand{\R}{\mathbb R}\renewcommand{\S}{\mathcal S}$This follows from Theorem IX.14, which states the following:

Let $T\in\S'(\R^n)$. Suppose that the Fourier transform $\hat T$ of $T$ can be continued analytically to the set $\{z\colon|\Im z|<A\}$ for some $A>0$. Suppose also that for each $B<A$ we have $\sup\{\|\hat T(\cdot+iy)\|_1\colon y\in(-B,B)\}<\infty$. Then $T$ is a bounded continuous function and for each $B<A$ there is a real number $C_B$ such that \begin{equation*} |T(x)|\le C_B e^{-B|x|} \end{equation*} for $x\in\R^n$.

Indeed, for the integral in question, \begin{equation*} I(t):=\int^\infty_0 \frac{\sin rt}{rt}\frac{r^4}{\sinh^2r} e^{-ar\coth(r)}\,dr, \end{equation*} and \begin{equation*} T(t):=t I(t), \tag{1}\label{1} \end{equation*} we have $T=\frac{\sqrt{2\pi}}{2i}\,\check f$, where \begin{equation*} f(r):=\frac{r^3}{\sinh^2r} e^{-ar\coth(r)} \end{equation*} and $\check f$ is the inverse Fourier transform of $f$ (defined by the formula $\check f(t):=\frac1{\sqrt{2\pi}}\,\int_\R e^{irt}f(r)\,dr$), so that \begin{equation*} \hat T=\frac{\sqrt{2\pi}}{2i}\,f. \end{equation*}

It is not hard to see that the function $T$ defined by \eqref{1} satisfies all the conditions of Theorem IX.14, cited above, with $A=\pi$. So, for each $B\in(0,\pi)$ there is a real number $C_B$ such that $|T(t)|\le C_B e^{-Bt}$ for all $t\ge0$ and hence \begin{equation*} |I(t)|\le C_B e^{-Bt} \end{equation*} for $t\ge1$. On the other hand, for all $t\in[0,1]$, \begin{equation*} |I(t)|\le c_a:=\int^\infty_0\frac{r^4}{\sinh^2r} e^{-ar\coth(r)}\,dr \le c_{a,B}\, e^{-Bt}, \end{equation*} where $c_{a,B}:=c_a e^B\in(0,\infty)$.

Thus, for each $B\in(0,\pi)$ and all $t\ge0$ \begin{equation*} |I(t)|\le\max(C_B,c_{a,B}) e^{-Bt}, \end{equation*} as desired.

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  • $\begingroup$ Good works@Iosif Pinelis. Thank you a lot. $\endgroup$ Jul 9, 2022 at 10:35
  • $\begingroup$ I m very gratufel for your help. Thanks a lot @Iosif Pinelis $\endgroup$ Jul 18, 2022 at 12:46
  • $\begingroup$ I voted on it five days ago Dear @ Iosif Pinelis. $\endgroup$ Jul 18, 2022 at 17:39
  • $\begingroup$ is it good now. I m sorry i don't know how todo this $\endgroup$ Jul 18, 2022 at 19:00
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    $\begingroup$ @zoranVicovic : Thank you for your appreciation. $\endgroup$ Jul 18, 2022 at 19:04

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